Answer: The value of y is 39.6 when x = 11 and the required expression is y(11) = 39.6
Step-by-step explanation:
Since we have given that
y varies directly as x.
If y = 18, when x = 5.
So, we need to find the value of y when x is 11.
According to question, we get that
![\dfrac{y}{x}=\dfrac{18}{5}=\dfrac{y}{11}\\\\18\times 11=5y\\\\198=5y\\\\\dfrac{198}{5}=y\\\\y=39.6](https://tex.z-dn.net/?f=%5Cdfrac%7By%7D%7Bx%7D%3D%5Cdfrac%7B18%7D%7B5%7D%3D%5Cdfrac%7By%7D%7B11%7D%5C%5C%5C%5C18%5Ctimes%2011%3D5y%5C%5C%5C%5C198%3D5y%5C%5C%5C%5C%5Cdfrac%7B198%7D%7B5%7D%3Dy%5C%5C%5C%5Cy%3D39.6)
Hence, the value of y is 39.6 when x = 11 and the required expression is y(11) = 39.6
Problem 1
<h3>Answer: False</h3>
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Explanation:
The notation (f o g)(x) means f( g(x) ). Here g(x) is the inner function.
So,
f(x) = x+1
f( g(x) ) = g(x) + 1 .... replace every x with g(x)
f( g(x) ) = 6x+1 ... plug in g(x) = 6x
(f o g)(x) = 6x+1
Now let's flip things around
g(x) = 6x
g( f(x) ) = 6*( f(x) ) .... replace every x with f(x)
g( f(x) ) = 6(x+1) .... plug in f(x) = x+1
g( f(x) ) = 6x+6
(g o f)(x) = 6x+6
This shows that (f o g)(x) = (g o f)(x) is a false equation for the given f(x) and g(x) functions.
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Problem 2
<h3>Answer: True</h3>
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Explanation:
Let's say that g(x) produced a number that wasn't in the domain of f(x). This would mean that f( g(x) ) would be undefined.
For example, let
f(x) = 1/(x+2)
g(x) = -2
The g(x) function will always produce the output -2 regardless of what the input x is. Feeding that -2 output into f(x) leads to 1/(x+2) = 1/(-2+2) = 1/0 which is undefined.
So it's important that the outputs of g(x) line up with the domain of f(x). Outputs of g(x) must be valid inputs of f(x).
3.2 is the Length of y 7.5 is the length of z and the length of x is 12.5
Answer:1206cm
Step-by-step explanation: